题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=4498
思路:分段积分。求出每个函数与y=100和每两个函数图像的交点,排序后枚举每段取最小值,分段积分。
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debu
using namespace std;
const int maxn=200;
const double eps=1e-8;
int best,n;
vector<double> x;
int a[maxn],b[maxn],k[maxn];
int A[maxn],B[maxn],C[maxn];
void make(int a,int b,int c)
{
if(a==0&&b==0) return ;
else if(a==0)
{
double tmp=-1.0*c/b;
if(tmp>=0&&tmp<=100) x.push_back(tmp);
}
else
{
double delta=b*b-4.0*a*c;
if(delta<0) return ;
else if(delta==0)
{
double tmp=-1.0*b/(2.0*a);
if(tmp<=100&&tmp>=0) x.push_back(tmp);
}
else
{
double tmp1=(-1.0*b+sqrt(delta))/(2.0*a);
if(tmp1<=100&&tmp1>=0) x.push_back(tmp1);
double tmp2=(-1.0*b-sqrt(delta))/(2.0*a);
if(tmp2<=100&&tmp2>=0) x.push_back(tmp2);
}
}
}
double F(double x)
{
return sqrt(1.0+(2.0*A[best]*x+1.0*B[best])*(2*A[best]*x+1.0*B[best]));
}
double simpson(double l,double r)
{
return (F(l)+4*F((l+r)/2.0)+F(r))*(r-l)/6.0;
}
double simpson(double l,double r,double all,double eps)
{
double m=(l+r)/2.0;
double L=simpson(l,m),R=simpson(m,r);
if(fabs(L+R-all)<=15*eps) return L+R+(L+R-all)/15;
return simpson(l,m,L,eps/2.0)+simpson(m,r,R,eps/2.0);
}
double simpson(double l,double r,double eps)
{
return simpson(l,r,simpson(l,r),eps);
}
int main()
{
#ifdef debug
freopen("in.in","r",stdin);
#endif // debug
int t;
scanf("%d",&t);
while(t--)
{
double ans=0.0;
x.clear();
scanf("%d",&n);
for(int i=1; i<=n; i++)
scanf("%d%d%d",&k[i],&a[i],&b[i]);
A[0]=B[0],C[0]=100.0;
for(int i=1; i<=n; i++)
{
A[i]=k[i];
B[i]=-2*k[i]*a[i];
C[i]=k[i]*a[i]*a[i]+b[i];
//make(A[i],B[i],C[i]);
}
for(int i=0;i<=n;i++)
for(int j=i+1;j<=n;j++)
make(A[i]-A[j],B[i]-B[j],C[i]-C[j]);
x.push_back(0),x.push_back(100);
sort(x.begin(),x.end());
/*for(int i=0;i<x.size()-1;i++)
cout<<x[i]<<" ";
cout<<x[x.size()-1]<<endl;*/
for(int i=0; i<x.size()-1; i++)
{
best=0;
double l=x[i],r=x[i+1];
if(fabs(l-r)<1e-8) continue;
double mid=(l+r)/2;
for(int j=1; j<=n; j++)
{
double tmp1=A[best]*mid*mid+B[best]*mid+C[best];
double tmp2=A[j]*mid*mid+B[j]*mid+C[j];
if(tmp2<tmp1) best=j;
}
//cout<<best<<" "<<l<<" "<<r<<" "<<simpson(l,r,eps)<<endl;
ans+=simpson(l,r,eps);
}
printf("%.2f\n",ans);
}
return 0;
}